Queue

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1299    Accepted Submission(s): 654 Problem Description N  people numbered from 1 to  N  are waiting in a bank for service. They all stand in a queue, but the queue never moves. It is lunch time now, so they decide to go out and have lunch first. When they get back, they don’t remember the exact order of the queue. Fortunately, there are some clues that may help. Every person has a unique height, and we denote the height of the  i -th person as  hi . The  i -th person remembers that there were  ki  people who stand before him and are taller than him. Ideally, this is enough to determine the original order of the queue uniquely. However, as they were waiting for too long, some of them get dizzy and counted  ki  in a wrong direction.  ki  could be either the number of taller people before or after the  i -th person. Can you help them to determine the original order of the queue?   Input The first line of input contains a number  T  indicating the number of test cases ( T≤1000 ). Each test case starts with a line containing an integer  N  indicating the number of people in the queue ( 1≤N≤100000 ). Each of the next  N  lines consists of two integers  hi  and  ki  as described above ( 1≤hi≤109,0≤ki≤N−1 ). Note that the order of the given  hi  and  ki  is randomly shuffled. The sum of  N  over all test cases will not exceed  106   Output For each test case, output a single line consisting of “Case #X: S”.  X  is the test case number starting from 1.  S  is people’s heights in the restored queue, separated by spaces. The solution may not be unique, so you only need to output the smallest one in lexicographical order. If it is impossible to restore the queue, you should output “impossible” instead.   Sample Input 3 3 10 1 20 1 30 0 3 10 0 20 1 30 0 3 10 0 20 0 30 1   Sample Output Case #1: 20 10 30 Case #2: 10 20 30 Case #3: impossible

题目大意：

已知有N个人，每个人的身高都不同，他们排成一列，而且他们记得他们的身前或者是身后有K个人比他高，问是否存在这样的可行序列，如果存在，输出最小字典序的解，否则输出impossible.

思路1：

①首先我们将序列按照身高从大到小排序，那么当前排列到第i个人的时候，已知所有之前排列上的人都比他高，所以只要将这个人插入到合适的位子即可，并且这个位子我们尽可能的靠向左边。

②当且仅当k>i-1的时候无解。

③那么过程用伸展树维护一下即可。

附上队长Ac代码：

#include <bits/stdc++.h> typedef long long int LL; typedef unsigned long long int LLu; using namespace std; const int N = 100000+5; const int MOD = 998244353; const int INF = 1e9+7; inline int read() { int x = 0,f=1; char ch = getchar(); for(; ch<'0'||'9'<ch; ch=getchar()) if(ch == '-') f=-1; for(; '0'<=ch&&ch<='9'; ch=getchar()) x=(x<<3)+(x<<1)+ch-'0'; return x*f; } #define lal puts("****"); #define pb push_back #define mp make_pair /********************************************************/ /*************SPLAY-tree************/ int n,m; int ch[N][2]; //ch[][0] lson ch[][1] rson int f[N]; //father int sz[N]; //size int val[N]; //value of node_i int cnt[N]; // counts of the node_i int root; //root of splay-tree int tot; //tot,total,is the number of node of tree void pushup(int x){ if(x)sz[x]=sz[ch[x][0]]+sz[ch[x][1]]+cnt[x]; } void rotate(int x,int k){ // k = 0 左旋， k = 1 右旋 int y=f[x];int z=f[y]; ch[y][!k]=ch[x][k];if(ch[x][k])f[ch[x][k]]=y; f[x]=z;if(z)ch[z][ch[z][1]==y]=x; f[y]=x;ch[x][k]=y; pushup(y),pushup(x); } void splay(int x,int goal){ for(int y=f[x];f[x]!=goal;y=f[x]) rotate(x,(ch[y][0]==x)); if(goal==0) root=x; } void newnode(int rt,int v,int fa){ // printf("newnode : rt = %d\n",rt); f[rt]=fa;val[rt]=v,sz[rt]=cnt[rt]=1; ch[rt][0]=ch[rt][1]=0; } void delnode(int &rt){ //其实是为内存回收做准备的 回头再完善 f[rt]=val[rt]=sz[rt]=cnt[rt]=0; ch[rt][0]=ch[rt][1]=rt=0; } /***************************以下是DEBUG***************************/ void Traversal(int rt){ if(!rt) return; Traversal(ch[rt][0]); printf("%d f[]=%d sz[]=%d lson=%d rson=%d val[]=%d\n",rt,f[rt],sz[rt],ch[rt][0],ch[rt][1],val[rt]); Traversal(ch[rt][1]); } void debug(){ printf("ROOT = %d <---\n",root); Traversal(root); } /**************************以下是前置操作**************************/ //以x为根的子树 的极值点 0 极小 1 极大 int extreme(int x,int k){ while(ch[x][k])x=ch[x][k];splay(x,0); return x; } //以x为根的子树 第k个数的位置 int kth(int x,int k){ if(sz[ch[x][0]]+1<=k&&k<=sz[ch[x][0]]+cnt[x]) return x; else if(sz[ch[x][0]]>=k) return kth(ch[x][0],k); else return kth(ch[x][1],k-sz[ch[x][0]]-cnt[x]); } int search(int rt,int x){ if(ch[rt][0]&&val[rt]>x) return search(ch[rt][0],x); else if(ch[rt][1]&&val[rt]<x)return search(ch[rt][1],x); else return rt; } /***************************以下是正经操作*************************/ //前驱 int prec(int x){ int k=search(root,x); splay(k,0);//debug(); if(val[k]<x) return k; return extreme(ch[k][0],1); } //后继 int sufc(int x){ int k=search(root,x); splay(k,0);//debug(); if(val[k]>x) return k; return extreme(ch[k][1],0); } int rk(int x){ int k=search(root,x); splay(k,0); return sz[ch[root][0]]+1; } //在第k个数后插入值为x的节点 void _insert(int k,int x){ int r=kth(root,k),rr=kth(root,k+1); splay(r,0),splay(rr,r); newnode(++tot,x,rr);ch[rr][0]=tot; for(r=rr;r;r=f[r])pushup(r); splay(rr,0); } /*****************************************************/ struct node { int h,k; }a[N]; bool cmp(node a,node b){ return a.h>b.h; } int ans[N]; void dfs(int rt){ if(!rt) return ; dfs(ch[rt][0]); ans[++ans[0]]=val[rt]; dfs(ch[rt][1]); } int main(){ int _ = 1,kcase = 0; for(scanf("%d",&_);_--;){ scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d%d",&a[i].h,&a[i].k); sort(a+1,a+n+1,cmp); printf("Case #%d:",++kcase); bool flag = false; for(int i=1;i<=n;i++) if(a[i].k>=i) flag =true; if(flag ){ puts(" impossible"); continue; } tot=0,root=1; newnode(++tot,-INF,0),newnode(++tot,INF,root); ch[root][1]=tot; for(int i=1;i<=n;i++){ a[i].k=min(a[i].k,i-1-a[i].k); _insert(a[i].k+1,a[i].h); } ans[0]=0; dfs(root); for(int i=2;i<ans[0];i++) printf(" %d",ans[i]); puts(""); } return 0; } 思路2：

①我们将序列按照身高从小到大排序，那么当前假设到第i个人入列的时候，我们之前入队的所有人都比他矮，那么他在如队的时候，只要身前留下min(k，n-i-k)个位子即可。

②所以我们一开始的时候，将一颗树状数组都置1，表示所有位子都还没有被占用过，过来一个人的时候，二分树上剩余位子的第min（k+1，n-i-k+1）个位子即可，如果存在这个位子，那么将这个人放到这个位子上，否则结果就是impossible.

③过程用树状数组套二分去跑就行了。

Ac代码：

#include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> using namespace std; struct node { int h,k; }a[100005]; int tree[100005];//树 int ans[100005]; int n; int cmp(node a,node b) { return a.h<b.h; } int lowbit(int x)//lowbit { return x&(-x); } int sum(int x)//求和求的是比当前数小的数字之和，至于这里如何实现，很简单：int sum=sum(a[i])； { int sum=0; while(x>0) { sum+=tree[x]; x-=lowbit(x); } return sum; } void add(int x,int c)//加数据。 { while(x<=n) { tree[x]+=c; x+=lowbit(x); } } int main() { int t; int kase=0; scanf("%d",&t); while(t--) { scanf("%d",&n); memset(tree,0,sizeof(tree)); for(int i=1;i<=n;i++)scanf("%d%d",&a[i].h,&a[i].k),add(i,1); sort(a+1,a+1+n,cmp); int flag=1; for(int i=1;i<=n;i++) { int pos=min(a[i].k+1,n-(i+a[i].k)+1); if(pos<1)flag=0; if(flag==0)break; int l=1; int r=n; int output=-1; while(r-l>=0) { int mid=(l+r)/2; if(sum(mid)<pos) { l=mid+1; } else { output=mid; r=mid-1; } } if(output==-1)flag=0; else ans[output]=a[i].h,add(output,-1); } printf("Case #%d: ",++kase); if(flag==0)printf("impossible\n"); else { for(int i=1;i<=n;i++) { if(i>1)printf(" "); printf("%d",ans[i]); } printf("\n"); } } }